We-know-that-vertex-form-of-parabola-is-given-as-y-a-x-h-2-k-From-the-given-diagram-of-bridge-that-resembles-a-parabola-we-have-a-vertex-points-of-0-30-and-other-points-due-to-towers-th

Question Number 174807 by Eulerian last updated on 11/Aug/22

We know that vertex form of parabola is given as

y = a {(x - h)}^{2} + k

From the given diagram of bridge that resembles a parabola,

we have a vertex points of (0, 30) and other points due to towers

that supports the parabolic - shape cable, which is (200, 150) .

∴ y = a {(x - 0)}^{2} + 30 = {ax}^{2} + 30

To find the value of^{'} a^{'}, {let}^{'} s use the given points other than vertex

150 = a {(200)}^{2} + 30

150 - 30 = 40000 a

a = \frac{120}{40, 000} = \frac{3}{1000}

∴ y = \frac{{3 x}^{2}}{1000} + 30

Also, since {we}^{'} re asked to find a function that gives a length of metal rod

needed relative to its distance from the midpoint of the bridge, with each rods

have an equal distance to each other, then we must consider another variable^{'} d^{'}

that represents the equal distance of metal rods relative to its decided quantity and

variable^{'} n^{'} given as positive integer that divides the distance of midpoint to tower .

d = \frac{200}{n} \Rightarrow nd = 200

Example :

Engineers decided to use 8 metal rodus, then we have d = \frac{200}{8} = 25

To calculate the length of each rods, {let}^{'} s use the formula above

First rod : y = \frac{3 {(0 \cdot 25)}^{2}}{1000} + 30 = 30 ft .

Second rod : y = \frac{3 {(1 \cdot 25)}^{2}}{1000} + 30 = 31.875 ft .

Third rod : y = \frac{3 {(2 \cdot 25)}^{2}}{1000} + 30 = 37.5 ft .

Fourth rod : y = \frac{3 {(3 \cdot 25)}^{2}}{1000} + 30 = 46.875 ft .